Question

A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are ${\gamma }_1$ and ${\gamma }_2$ respectively. If their temperature are increased by $\Delta T$ each, the fraction of the volume of metal submerged in mercury changes by a factor of

• A. $({\gamma }_2-{\gamma }_1)\Delta T$
• B. ${\gamma }_1\Delta T$
• C. $(1+({\gamma }_2-{\gamma }_1\left)\Delta T\right)$
• D. $\frac{{\gamma }_2}{{\gamma }_1}$

Answer 1

The correct option is A $({\gamma }_2-{\gamma }_1)\Delta T$

When the temperature is increased by a $\Delta T$
New volume of metal will be $\left(V_1\right)=V_0(1+{\gamma }_1\Delta T).....\left(1\right)$
New volume of mercury will be $\left(V_2\right)=V_0(1+{\gamma }_2\Delta T).....\left(2\right)$
From $\left(1\right)\text{and}\left(2\right)$, we get
$\frac{V_2}{V_1}=\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$
$\Rightarrow \frac{V_2}{V_1}=(1+{\gamma }_2\Delta T){(1+{\gamma }_1\Delta T)}^{-1}$
Applying binomial expansion,
$\frac{V_2}{V_1}=(1+{\gamma }_2\Delta T)(1-{\gamma }_1\Delta T)$
$\Rightarrow \frac{V_2}{V_1}=(1+{\gamma }_2\Delta T-{\gamma }_1\Delta T-{\gamma }_1{\gamma }_2(\Delta T{)}^2)$
$\Rightarrow \frac{V_2}{V_1}-1={\gamma }_2\Delta T-{\gamma }_1\Delta T$
[neglecting second order terms]
$\Rightarrow \frac{\Delta V}{V_1}=({\gamma }_2\Delta T-{\gamma }_1\Delta T)$
[$V_2-V_1=\Delta V$]
$\therefore$ The fraction of the volume of metal submerged in mercury changes by a factor of $({\gamma }_2-{\gamma }_1)\Delta T$
Hence, option (a) is the correct answer.